Given that $E_1$ and $E_2$ are independent events and that $E_1 \, \vert \, S$ and $E_2\, \vert \, S$ are independent events, is it possible to show that $$p\left(E_1 \cap E_2\, \middle\vert \, S \right)=p\left(E_1\, \middle\vert \, S \right)p\left(E_2\, \middle\vert \, S \right)\text{.}$$ By the definition of independence of events, $p\left(E_1\, \middle\vert \, S \right)p\left(E_2\, \middle\vert \, S \right)=p\bigl(\left(E_1 \, \middle\vert \, S\right) \cap \left(E_2 \, \middle\vert \, S\right)\bigr)$, which gives $$p\left(E_1 \cap E_2\, \middle\vert \, S \right)=p\bigl(\left(E_1 \, \middle\vert \, S\right) \cap \left(E_2 \, \middle\vert \, S\right)\bigr)\text{.}$$

This relation seems reasonable for independent $E_1$ and $E_2$ (so that $p\left(E_1 \cap E_2\right)=p\left(E_1\right)p\left(E_2\right)$), but I can't figure out how to construct an explicit proof for it.

Adjunct #1: If it is not possible to prove this relationship using the aforementioned assumptions, can it be proven if we assume $E_1$, $E_2$, and $S$ are pairwise independent? Mutually independent? I do not believe that either of these assumptions are valid, nor that they are necessary to complete the proof.

Adjunct #2: I've seen that the relationship $p\left(E_1 \cap E_2\, \middle\vert \, S \right)=p\left(E_1\, \middle\vert \, S \right)p\left(E_2\, \middle\vert \, S \right)$ is characteristic of conditional independence, but I'm not sure exactly what this means or if it can be used to solve the problem.

Edit: Hopefully this will help—I'm including the question in its entirety, which comes from Rosen's Discrete Mathematics and its Applications, 8th Edition. Question 23 in Chapter 7.3 concerns Bayes' theorem as it is applied to spam filters. Rosen asks:

Suppose that $E_1$ and $E_2$ are the events that an incoming mail message contains the words $w_1$ and $w_2$, respectively. Assuming that $E_1$ and $E_2$ are independent events and that $E_1\, \vert\, S$ and $E_2\, \vert\, S$ are independent events, where $S$ is the event that an incoming message is spam, and that we have no prior knowledge regarding whether or not the message is spam, show that $$p\left(S \, \middle\vert \, E_1 \cap E_2 \right)=\frac{p\left(E_1\, \middle\vert \, S \right)p\left(E_2\, \middle\vert \, S \right)}{p\left(E_1\, \middle\vert \, S \right)p\left(E_2\, \middle\vert \, S \right)+p\left(E_1\, \middle\vert \, \bar{S} \right)p\left(E_2\, \middle\vert \, \bar{S} \right)}\text{.}$$

From the solution manual:

First, by Bayes' theorem, we have $$\left(S \, \middle\vert \, E_1 \cap E_2 \right)=\frac{p\left(E_1 \cap E_2 \, \middle\vert \, S\right)p\left(S \right)}{p\left(E_1 \cap E_2 \, \middle\vert \, S\right)p\left(S \right)+p\left(E_1 \cap E_2 \, \middle\vert \,\bar{S}\right)p\left(\bar{S} \right)}\text{.}$$ Next, because we are assuming no prior knowledge about whether a message is or is not spam, we set $p(S) = p\left(\bar{S} \right) = 0.5$, and so the equation above simplifies to $$\left(S \, \middle\vert \, E_1 \cap E_2 \right)=\frac{p\left(E_1 \cap E_2 \, \middle\vert \, S\right)}{p\left(E_1 \cap E_2 \, \middle\vert \, S\right)+p\left(E_1 \cap E_2 \, \middle\vert \, \bar{S}\right)}\text{.}$$ Finally, because of the assumed independence of $E_1$, $E_2$, and $S$, we have $p\left(E_1 \cap E_2\, \middle\vert \, S \right)=p\left(E_1\, \middle\vert \, S \right)p\left(E_2\, \middle\vert \, S \right)$, and similarly for the $\bar{S}$. Thus the equation is equivalent to what we were asked to show.

  • $\begingroup$ What do you mean by "$\ E_i|S\ $"? The expression $\ P\left(E_i|S\right)\ $ is used to represent the conditional probability of $\ E_i\ $ given $\ S\ $, defined by $$ P(E_i|S)\stackrel{\text{def}}{=}\frac{P(E_i\cap S)}{P(S)}\ $$ but, as far as I know, there is no "event" commonly represented by the expression "$\ E_i|S\ $". Two events $\ E_1\ $ and $\ E_2\ $ are *defined* to be conditionally independent given $\ S\ $ if $$ P(E_1\cap E_2|S)=P(E_1|S)P(E_2|S)\ . $$ $\endgroup$ Commented Apr 8, 2020 at 4:23
  • $\begingroup$ To be frank, I'm not entirely sure what is meant by that either. This seems to be the only instance of the text using the conditional probability notation outside of a probability expression, and I can't find anything online that offers any further explanation. If it helps, I've now included the full problem in my question, as well as its corresponding solution from the solutions manual. My confusion stems from the final step of the solution, as well as the conditional probability notation that you mentioned. $\endgroup$
    – RyanC
    Commented Apr 11, 2020 at 7:44

1 Answer 1


$E|S$ without context is an abuse of notation; it is not an event, as your book erroneously says. The incorrect phrase "$E1|S$ and $E2|S$ are independent events" can only mean that events $E1$ and $E2$ are conditionally independent given $S$, that is $$P(E_1\cap E_2|S)=P(E_1|S)P(E_2|S)$$ Since this is a definition (of conditional independence) there is nothing to prove here.

  • $\begingroup$ Proof $\endgroup$
    – HeyJude
    Commented Nov 9, 2023 at 21:39

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